Control Systems

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

Page 128

UNIT 6

The state diagram of a system is shown below. A system is o = AX + Bu ; described by the state-variable equations X y = CX + Du

CONTROL SYSTEMS

2013 6.1

ONE MARK

6.4

The Bode plot of a transfer function G ^s h is shown in the figure below.

6.5

The state-variable equations of the system shown in the figure above are o = >- 1 0 H X + >- 1H u o = >- 1 0 H X + >- 1H u X X (B) (A) -1 -1 1 -1 1 1 y = 61 - 1@ X + u y = 6- 1 - 1@ X + u o = >- 1 0 H X + >- 1H u o = >- 1 - 1H X + >- 1H u X X (C) (D) -1 -1 1 0 -1 1 y = 6- 1 - 1@ X - u y = 61 - 1@ X - u The state transition matrix eAt of the system shown in the figure above is e-t 0 e-t 0 (A) > -t -tH (B) > H -t te e - te e-t e-t 0 (C) > -t -tH e e

The gain _20 log G ^s h i is 32 dB 10 rad/s respectively. The phase (A) 39.8 s (C) 32 s

and - 8 dB at 1 rad/s and is negative for all w. Then G ^s h is (B) 392.8 s (D) 322 s TWO MARKS

s+1 5s2 + 6s + 2 (C) 2 s + 1 s + 4s + 2

(B) 5 (D) 100

2012 6.7

6.8

s+1 s2 + 6s + 2 (D) 2 1 5s + 6s + 2

(B) K = 3 and a = 0.75 (D) K = 2 and a = 0.5

The state variable description of an LTI system is given by Jxo1N J 0 a1 0NJx1N J0N K O K OK O K O Kxo2O = K 0 0 a2OKx2O + K0O u Kxo O Ka 0 0OKx 3O K 1O 3 3 L P L PL P L P Jx1N K O y = _1 0 0iKx2O Kx 3O L P where y is the output and u is the input. The system is controllable for (A) a1 ! 0, a2 = 0, a 3 ! 0 (B) a1 = 0, a2 ! 0, a 3 ! 0 (C) a1 = 0, a 3 ! 0, a 3 = 0 (D) a1 ! 0, a2 ! 0, a 3 = 0 2011

6.9

TWO MARKS

The feedback system shown below oscillates at 2 rad/s when

(A) K = 2 and a = 0.75 (C) K = 4 and a = 0.5

(B)

Statement for Linked Answer Questions 4 and 5:

A system with transfer function G (s) =

.n

The signal flow graph for a system is given below. The transfer Y ^s h function for this system is U ^s h

(A)

ONE MARK

(s2 + 9) (s + 2) (s + 1) (s + 3) (s + 4) is excited by sin (wt). The steady-state output of the system is zero at .in w = 1 rad/s (B) w = 2 rad/s o(A) c . a i (D) w = 4 rad/s od (C) w = 3 rad/s 6.6

The open-loop transfer function of a dc motor is givenww as w w ^s h = 10 . When connected in feedback as shown below, the Va ^s h 1 + 10s approximate value of Ka that will reduce the time constant of the closed loop system by one hundred times as compared to that of the open-loop system is

(A) 1 (C) 10 6.3

2012

O N

2013 6.2

A I D

e-t - te-t (D) > H 0 e-t

ONE MARK

The root locus plot for a system is given below. The open loop transfer function corresponding to this plot is given by

GATE Electronics & Communication by RK Kanodia (Now in 3 Volumes) Purchase Online from our portal www.nodia.co.in and get maximum available discount

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

Page 129

100 . s (s + 10) 2 The plant is placed in a unity negative feedback configuration as shown in the figure below. The input-output transfer function of a plant H (s) =

s (s + 1) (s + 2) (s + 3) (s + 1) (B) G ^s h H ^s h = k s (s + 2) (s + 3) 2 1 (C) G ^s h H ^s h = k s (s - 1) (s + 2) (s + 3) (s + 1) (D) G ^s h H ^s h = k s (s + 2) (s + 3) 6.10

The gain margin of the system under closed loop unity negative feedback is (A) 0 dB (B) 20 dB (C) 26 dB (D) 46 dB

6.12

(A) G ^s h H ^s h = k

The signal flow graph that DOES NOT model the plant transfer function H (s) is

6.13

For the transfer function G (jw) = 5 + jw , the corresponding Nyquist plot for positive frequency has the form

A I D

O N

2010

. The transfer function Y (s) /R (s) of the system shown is

ia od

6.14

ONE MARK

in co.

n

. ww

w

2011 6.11

TWO MARKS

The block diagram of a system with one input u and two outputs y1 and y2 is given below.

6.15

A state space model of the above system in terms of the state vector x and the output vector y = [y1 y2]T is (A) xo = [2] x + [1] u ; y = [1 2] x 1 (B) xo = [- 2] x + [1] u; y = > H x 2 -2 0 1 (C) xo = > x + > H u ; y = 81 2B x H 0 -2 1 2 0 1 1 (D) xo = > H x + > H u ; y = > H x 0 2 1 2

6.16

1 s+1 (C) 2 (D) 2 s+1 s+3 Y (s) A system with transfer function has an output = s s+p X ( s ) p y (t) = cos a2t - k 3 for the input signal x (t) = p cos a2t - p k. Then, the system param2 eter p is (B) 2/ 3 (A) 3 (A) 0

(B)

(C) 1

(D)

3 /2

For the asymptotic Bode magnitude plot shown below, the system transfer function can be

Common Data For Q. 7.4 & 7.5

GATE GUIDE and GATE CLOUD by RK Kanodia & Ashish Murolia GATE GUIDEs and CLOUDs are subjectwise books exclusive for GATE EC Examination. Download a sample chapter at

www.nodia.co.in

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

(A) 10s + 1 0.1s + 1 (C) 100s 10s + 1

Page 130

(B) 100s + 1 0.1s + 1 (D) 0.1s + 1 10s + 1

(C) The system is uncontrollable for all values of p and q (D) We cannot conclude about controllability from the given data 2009

2010 6.17

TWO MARKS

6.22

TWO MARKS

The feedback configuration and the pole-zero locations of 2 G (s) = s2 - 2s + 2 s + 2s + 2 are shown below. The root locus for negative values of k , i.e. for - 3 < k < 0 , has breakaway/break-in points and angle of departure at pole P (with respect to the positive real axis) equal to

A unity negative feedback closed loop system has a plant with the and a controller Gc (s) in the transfer function G (s) = 2 1 s + 2s + 2 feed forward path. For a unit set input, the transfer function of the controller that gives minimum steady state error is (B) Gc (s) = s + 2 (A) Gc (s) = s + 1 s+2 s+1 (s + 1) (s + 4) (C) Gc (s) = (D) Gc (s) = 1 + 2 + 3s s (s + 2) (s + 3)

Common Data For Q. 7.10 & 7.11 : The signal flow graph of a system is shown below:

6.23

6.18

6.19

The state variable representation of the system can be -1 1 0 1 1 0 o= > xo = > x +> Hu H x x + u H > H (A) (B) -1 0 2 -1 0 2 yo = 80 0.5B x yo = [0 0.5] x 1 1 0 xo = > x +> Hu H (C) (D) -1 0 2 yo = 80.5 0.5B x The transfer function of the system is (B) (A) s2+ 1 s +1 (C) 2 s + 1 (D) s +s+1 2009

6.20

-1 xo = > -1 yo = 80.5

(C) ! 3 and 0c

(D) ! 3 and 45c

The unit step response of an under-damped second order system has steady state value of -2. Which one of the following transfer functions has theses properties ? (A) 2 - 2.24 (B) 2 - 3.82 s + 2.59s + 1.12 s + 1.91s + 1.91 - 382 (C) 2 - 2.24 (D) 2 s - 2.59s + 1.12 s - 1.91s + 1.91

O N

Common Data For Q. 7.16 and 7.17 :

no w.

The .in Nyquist plot of a stable transfer function G (s) is shown in the ofigure c . are interested in the stability of the closed loop system in ia

d

the feedback configuration shown.

ww

ONE MARK

The magnitude plot of a rational transfer function G (s) with real coefficients is shown below. Which of the following compensators has such a magnitude plot ?

6.24

6.25

(A) Lead compensator (C) PID compensator 6.21

(B) ! 2 and 45c

A I D

1 0 x +> Hu H 0 2 0.5B x

s-1 s2+1 s-1 s2+s+1

(A) ! 2 and 0c

(B) Lag compensator (D) Lead-lag compensator

Consider the system dx = Ax + Bu with A = =1 0G and B = = p G q 0 1 dt where p and q are arbitrary real numbers. Which of the following statements about the controllability of the system is true ? (A) The system is completely state controllable for any nonzero values of p and q (B) Only p = 0 and q = 0 result in controllability

Which of the following statements is true ? (A) G (s) is an all-pass filter (B) G (s) has a zero in the right-half plane (C) G (s) is the impedance of a passive network (D) G (s) is marginally stable The gain and phase margins of G (s) for closed loop stability are (A) 6 dB and 180c (B) 3 dB and 180c (C) 6 dB and 90c (D) 3 dB and 90c 2008

6.26

ONE MARKS

Step responses of a set of three second-order underdamped systems all have the same percentage overshoot. Which of the following diagrams represents the poles of the three systems ?

GATE Electronics & Communication by RK Kanodia (Now in 3 Volumes) Purchase Online from our portal www.nodia.co.in and get maximum available discount

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

Page 131

6.30

Jx1N R b - g 0 VJx1N R0 0 V W u1 WK O S K O S (A) d K x2O = S g a 0 WK x2O+ S0 1 We o u2 dt K O S x3 S- a b 0 WWK x3O SS1 0 WW L P L P Jx1N RT0 a g XVJx1N TR1 0 XV W u1 WK O S K O S (B) d K x2O = S0 - a - g WK x2O+ S0 1 We o u2 dt K O S x3 S0 b - b WWK x3O SS0 0 WW L P TR L P Jx1N - a b 0 VXJx1N RT1 0 VX W u1 W S K O K O S (C) d K x2O = S- b - g 0 WK x2O+ S0 1 We o u2 dt K O S x3 S a g 0 WWK x3O SS0 0 WW L P TR L P Jx1N - a 0 b XVJx1N TR1 0XV W u1 WK O S K O S (D) d K x2O = S g 0 a WK x2O+ S0 1 We o u2 dt K O S x3 S- b 0 - a WWK x3O SS0 0 WW L P T X T XL P function A certain system has transfer G (s) = 2 s + 8 s + as - 4 where a is a parameter. Consider the standard negative unity feedback configuration as shown below

6.27

The pole-zero given below correspond to a

(A) Law pass filter (C) Band filter 2008 6.28

Which of the following statements is true? (A) The closed loop systems is never stable for any value of a (B) For some positive value of a, the closed loop system is stable, but not for all positive values. (C) For all positive values of a, the closed loop system is stable. (D) The closed loop system stable for all values of a, both positive and negative.

(B) High pass filter (D) Notch filter

A I D

TWO MARKS

Group I lists a set of four transfer functions. Group II gives a list of possible step response y (t). Match the step responses with the corresponding transfer functions.

O N

no w.

d

ww

6.32

6.33

(A) P - 3, Q - 1, R - 4, S - 2 (C) P - 2, Q - 1, R - 4, S - 2 6.29

The .innumber of open right half plane of 10 is G (s) = 5 4 3 s + 2s + 3s + 6s2 + 5s + 3 (A) 0 (B) 1 (C) 2 (D) 3

co ia.

6.31

The magnitude of frequency responses of an underdamped second order system is 5 at 0 rad/sec and peaks to 10 at 5 2 rad/sec. 3 The transfer function of the system is 500 (A) 2 (B) 2 375 s + 10s + 100 s + 5s + 75 720 (C) 2 (D) 2 1125 s + 12s + 144 s + 25s + 225 Group I gives two possible choices for the impedance Z in the diagram. The circuit elements in Z satisfy the conditions R2 C2 > R1 C1. The transfer functions V0 represents a kind of controller. Vi

(B) P - 3, Q - 2, R - 4, S - 1 (D) P - 3, Q - 4, R - 1, S - 2 Match the impedances in Group I with the type of controllers in Group II

A signal flow graph of a system is given below

The set of equalities that corresponds to this signal flow graph is

GATE GUIDE and GATE CLOUD by RK Kanodia & Ashish Murolia GATE GUIDEs and CLOUDs are subjectwise books exclusive for GATE EC Examination. Download a sample chapter at

www.nodia.co.in

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

(A) Q - 1, R - 2 (C) Q - 2, R - 3

(B) Q - 1, R - 3 (D) Q - 3, R - 2

Page 132 6.40

The state space representation of a separately excited DC servo motor dynamics is given as dw dt

> di H = =- 1 - 10G=ia G + =10Gu dt -1

1 w

0

o

2007 6.34

ONE MARK

2007 6.35

6.37

6.38

10 s2 + 11s + 11 (C) 2 10s + 10 s + 11s + 11

(A)

Statement for linked Answer Question 8.33 & 8.34 : Consider a linear system whose state space representation is 1 x (t) = Ax (t). If the initial state vector of the system is x (0) = = G, -2 e-2x then the system response is x (t) = > H. If the itial state vector - 2e-2t 1 of the system changes to x (0) = = G, then the system response -2 e-t becomes x (t) = > -tH -e

(B) KP = 100, KD = 0.9 (D) KP = 10, KD = 0.9

The transfer function of a plant is 6.41 The eigenvalue and eigenvector pairs (li vi) for the system are 5 1 1 1 1 T (s) = (A) e- 1 = Go and e- 2 = Go (B) e- 1, = Go and e2, = Go (s + 5)( s2 + s + 1) -1 -2 -1 -2 The second-order approximation of T (s) using dominant pole con1 1 1 1 cept is (C) e- 1, = Go and e- 2, = Go (D) e- 2 = Go and e1, = Go -1 -2 -1 -2 1 5 (A) (B) 6.42 The system matrix A is (s + 5)( s + 1) (s + 5)( s + 1) 1 1 .in 0 1 o(A) c (C) 2 5 (D) 2 1 (B) = =- 1 1G . a - 1 - 2G i s +s+1 s +s+1 d o 2 1 0 1 The open-loop transfer function of a plant is given as G (s) = s 1-w ..n 1 (C) = (D) = G w -1 -1 - 2 - 3G If the plant is operated in a unity feedback configuration, then w the lead compensator that an stabilize this control system is 10 (s - 1) 10 (s + 4) 2006 ONE MARK (B) (A) s+2 s+2 6.43 The open-loop function of a unity-gain feedback control system is 10 (s + 2) 2 (s + 2) (C) (D) given by s + 10 s + 10 K G (s) = (s + 1)( s + 2) A unity feedback control system has an open-loop transfer function K The gain margin of the system in dB is given by G (s) = s (s2 + 7s + 12) (A) 0 (B) 1

A I D

O N 2

The gain K for which s = 1 + j1 will lie on the root locus of this system is (A) 4 (B) 5.5 (C) 6.5 (D) 10 6.39

1 s2 + 11s + 11 (D) 2 1 s + s + 11 (B)

TWO MARKS

A control system with PD controller is shown in the figure. If the velocity error constant KV = 1000 and the damping ratio z = 0.5 , then the value of KP and KD are

(A) KP = 100, KD = 0.09 (C) KP = 10, KD = 0.09 6.36

where w is the speed of the motor, ia is the armature current and w (s) of the motor u is the armature voltage. The transfer function U (s) is

If the closed-loop transfer function of a control system is given as s-5 , then It is T (s) (s + 2)( s + 3) (A) an unstable system (B) an uncontrollable system (C) a minimum phase system (D) a non-minimum phase system

The asymptotic Bode plot of a transfer function is as shown in the figure. The transfer function G (s) corresponding to this Bode plot is

(C) 20 2006 6.44

6.45

1 (s + 1)( s + 20) 100 (C) s (s + 1)( s + 20)

1 s (s + 1)( s + 20) 100 (D) s (s + 1)( 1 + 0.05s) (B)

6.46

TWO MARKS

Consider two transfer functions G1 (s) = 2 1 and s + as + b . G2 (s) = 2 s s + as + b The 3-dB bandwidths of their frequency responses are, respectively (B) a2 + 4b , a2 - 4b (A) a2 - 4b , a2 + 4b (C)

(A)

(D) 3

a2 - 4b , a2 - 4b

(D)

a 2 + 4b , a 2 + 4b

The Nyquist plot of G (jw) H (jw)for a closed loop control system, passes through (- 1, j0) point in the GH plane. The gain margin of the system in dB is equal to (A) infinite (B) greater than zero (C) less than zero (D) zero The positive values of K and a so that the system shown in the figures below oscillates at a frequency of 2 rad/sec respectively are

GATE Electronics & Communication by RK Kanodia (Now in 3 Volumes) Purchase Online from our portal www.nodia.co.in and get maximum available discount

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

(A) 1, 0.75 (C) 1, 1 6.47

6.48

Page 133

(B) 2, 0.75 (D) 2, 2

The transfer function of a phase lead compensator is given by Gc (s) = 1 + 3Ts where T > 0 The maximum phase shift provide by 1 + Ts such a compensator is (A) p (B) p 2 3 (C) p (D) p 4 6 A linear system is described by the following state equation 0 1 Xo (t) = AX (t) + BU (t), A = = - 1 0G The state transition matrix of the system is cos t sin t - cos t sin t (A) = (B) = G - sin t cos t - sin t - cos t G - cos t - sin t cos t - sin t (C) = (D) = G - sin t cos t cos t sin t G

2005

Consider a unity - gain feedback control system whose open - loop 1 transfer function is : G (s) = as + s2 The value of a so that the system has a phase - margin equal to p 4 is approximately equal to (A) 2.40 (B) 1.40 (C) 0.84 (D) 0.74

O N

A I D .in

no w.

6.50

6.51

ONE MARK

6.55

A linear system is equivalently represented by two sets of state equations : Xo = AX + BU and Wo = CW + DU The eigenvalues of the representations are also computed as [l] and [m]. Which one of the following statements is true ? (A) [l] = [m] and X = W (B) [l] = [m] and X ! W (C) [l] ! [m] and X = W (D) [l] = [m] and X ! W

6.52

Which one of the following polar diagrams corresponds to a lag network ?

co ia.

d

w With the value of a set for a phase - margin of p , the value ofw unit 4 - impulse response of the open - loop system at t = 1 second is equal to (A) 3.40 (B) 2.40 (C) 1.84 (D) 1.74 2005

TWO MARKS

The polar diagram of a conditionally stable system for open loop gain K = 1 is shown in the figure. The open loop transfer function of the system is known to be stable. The closed loop system is stable for

6.54

Statement for Linked Answer Questions 7.41 & 7.42 :

6.49

Despite the presence of negative feedback, control systems still have problems of instability because the (A) Components used have non- linearities (B) Dynamic equations of the subsystem are not known exactly. (C) Mathematical analysis involves approximations. (D) System has large negative phase angle at high frequencies.

6.53

6.56

6.57

(A) K < 5 and 1 < K < 1 (B) K < 1 and 1 < K < 5 2 8 8 2 (C) K < 1 and 5 < K (D) K > 1 and 5 > K 8 8 In the derivation of expression for peak percent overshoot - px Mp = exp e o # 100% 1 - x2 Which one of the following conditions is NOT required ? (A) System is linear and time invariant (B) The system transfer function has a pair of complex conjugate poles and no zeroes. (C) There is no transportation delay in the system. (D) The system has zero initial conditions. A ramp input applied to an unity feedback system results in 5% steady state error. The type number and zero frequency gain of the system are respectively (A) 1 and 20 (B) 0 and 20 (C) 0 and 1 (D) 1 and 1 20 20 2 A double integrator plant G (s) = K/s , H (s) = 1 is to be compensated to achieve the damping ratio z = 0.5 and an undamped natural frequency, wn = 5 rad/sec which one of the following compensator Ge (s) will be suitable ? (B) s + 99 (A) s + 3 s + 99 s+3

GATE GUIDE and GATE CLOUD by RK Kanodia & Ashish Murolia GATE GUIDEs and CLOUDs are subjectwise books exclusive for GATE EC Examination. Download a sample chapter at

www.nodia.co.in

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

(C) 6.58

s-6 s + 8.33

Page 134

(D) s - 6 s

excited with 10u (t). The time at which the output reaches 99% of its steady state value is (A) 2.7 sec (B) 2.5 sec (C) 2.3 sec (D) 2.1 sec

K (1 - s) An unity feedback system is given as G (s) = . s (s + 3) Indicate the correct root locus diagram. 6.65

6.66

A system has poles at 0.1 Hz, 1 Hz and 80 Hz; zeros at 5 Hz, 100 Hz and 200 Hz. The approximate phase of the system response at 20 Hz is (A) - 90c (B) 0c (C) 90c (D) - 180c Consider the signal flow graph shown in Fig. The gain x5 is x1

(A)

Statement for Linked Answer Question 40 and 41 :

6.59

6.60

The open loop transfer function of a unity feedback system is given by -2s G (s) = 3e s (s + 2) The gain and phase crossover frequencies in rad/sec are, respectively (A) 0.632 and 1.26 (B) 0.632 and 0.485 (C) 0.485 and 0.632 (D) 1.26 and 0.632

6.61

6.62

6.67

.in open-loop transfer function of a unity feedback system is oThe c . ia

6.68

no w.

d

The gain margin for the system with open-loop transfer function ww 2 (1 + s) , is G (s) H (s) = s2 (A) 3 (B) 0 (C) 1 (D) - 3

K Given G (s) H (s) = .The point of intersection of the s (s + 1)( s + 3) asymptotes of the root loci with the real axis is (A) - 4 (B) 1.33 (C) - 1.33 (D) 4

6.69

6.70

2004 6.63

TWO MARKS

Consider the Bode magnitude plot shown in the fig. The transfer function H (s) is

6.71

(A)

(s + 10) (s + 1)( s + 100)

(B)

10 (s + 1) (s + 10)( s + 100)

102 (s + 1) 103 (s + 100) (D) (s + 10)( s + 100) (s + 1)( s + 10) A causal system having the transfer function H (s) = 1/ (s + 2) is

K s (s2 + s + 2)( s + 3) The range of K for which the system is stable is (A) 21 > K > 0 (B) 13 > K > 0 4 (C) 21 < K < 3 (D) - 6 < K < 3 4 For the polynomial P (s) = s2 + s 4 + 2s3 + 2s2 + 3s + 15 the number of roots which lie in the right half of the s -plane is (A) 4 (B) 2 (C) 3 (D) 1

6.72

G (s) =

The state variable equations of a system are : xo1 =- 3x1 - x2 = u, xo2 = 2x1 and y = x1 + u . The system is (A) controllable but not observable (B) observable but not controllable (C) neither controllable nor observable (D) controllable and observable Given A = =

1 0 , the state transition matrix eAt is given by 0 1G

0 e-t (A) > -t H e 0

et 0 (B) = G 0 et

e-t 0 (C) > H 0 e-t

0 et (D) = t G e 0

2003

(C) 6.64

bedg 1 - (be + cf + dg) 1 - (be + cf + dg) + bedg (D) abcd

(B)

sin (- 4t) + 2 sin (- t) - 2 sin (- 4t) + 2 sin (- t) (A) 1 = G 3 - sin (- 4t) + sin (- t) 2 sin (- 4t) + sin (- t) sin (- 2t) sin (2t) (B) = sin (t) sin (- 3t)G sin (4t) + 2 sin (t) 2 sin (- 4t) - 2 sin (- t) (C) 1 = 2 sin (4t) + sin (t) G 3 - sin (- 4t) + sin (t) cos (- t) + 2 cos (t) 2 cos (- 4t) + 2 cos (- t) (D) 1 = G 3 - cos (- 4t) + cos (- t) - 2 cos (- 4t) + cos (t)

A I D

O N ONE MARK

abcd 1 - (be + cf + dg) + bedg -2 2 If A = = , then sin At is 1 - 3G (C)

Based on the above results, the gain and phase margins of the system will be (A) -7.09 dB and 87.5c (B) 7.09 dB and 87.5c (C) 7.09 dB and - 87.5c (D) - 7.09 and - 87.5c 2004

1 - (be + cf + dg) abcd

ONE MARK

Fig. shows the Nyquist plot of the open-loop transfer function G (s) H (s) of a system. If G (s) H (s) has one right-hand pole, the

GATE Electronics & Communication by RK Kanodia (Now in 3 Volumes) Purchase Online from our portal www.nodia.co.in and get maximum available discount

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

Page 135

closed-loop system is

(A) always stable (B) unstable with one closed-loop right hand pole (C) unstable with two closed-loop right hand poles (D) unstable with three closed-loop right hand poles 6.73

A PD controller is used to compensate a system. Compared to the uncompensated system, the compensated system has (A) a higher type number (B) reduced damping (C) higher noise amplification (D) larger transient overshoot 6.78

2003 6.74

TWO MARKS

The signal flow graph of a system is shown in Fig. below. The transfer function C (s)/ R (s) of the system is 6.79

(A)

6.76

(B)

s (s + 2) s + 29s + 6

(D)

A I D

6s 2 s + 29s + 6

tet (A) = G t et (C) = t G te

2

2

O N

6.82

(s + 0.1) 3 (s + 0.1) 3 7 (B) 10 (s + 10)( s + 100) (s + 10) 2 (s + 100) 2 (s + 0.1) (s + 0.1) 3 (C) (D) (s + 10) 2 (s + 100) (s + 10)( s + 100) 2 A second-order system has the transfer function C (s) = 2 4 R (s) s + 4s + 4 With r (t) as the unit-step function, the response c (t) of the system is represented by (A) 108

6.77

The zero-input response of a system given by the state-space equation 1 0 x1 x1 (0) 1 xo1 =xo G = =1 1G=x G and =x (0)G = = 0 G is 2 2 2 et (B) = G t t (D) = t G te

s (s + 27) s + 29s + 6 K .in The root locus of system G (s) H (s) = has the breako c 2002 ONE MARK . s (s + 2)( s + 3) ia away point located at d o .n6.80 (A) (- 0.5, 0) (B) (- 2.548, 0) Consider a system with transfer function G (s) = 2s + 6 . Its w ks + s + 6 ww damping ratio will be 0.5 when the value of k is (C) (- 4, 0) (D) (- 0.784, 0) (A) 2 (B) 3 6 The approximate Bode magnitude plot of a minimum phase system is shown in Fig. below. The transfer function of the system is (C) 1 (D) 6 6 6.81 Which of the following points is NOT on the root locus of a system k with the open-loop transfer function G (s) H (s) = s (s + 1)( s + 3) (A) s =- j 3 (B) s =- 1.5 (C) s =- 3 (D) s =- 3 (C)

6.75

6 2 s + 29s + 6

The gain margin and the phase margin of feedback system with 8 are G (s) H (s) = (s + 100) 3 (A) dB, 0c (B) 3, 3 (C) 3, 0c (D) 88.5 dB, 3

6.83

The phase margin of a system with the open - loop transfer function (1 - s) G (s) H (s) = (1 + s)( 2 + s) (A) 0c (B) 63.4c (C) 90c (D) 3 The transfer function Y (s)/ U (s) of system described by the state equation xo (t) =- 2x (t) + 2u (t) and y (t) = 0.5x (t) is 1 (A) 0.5 (B) (s - 2) (s - 2) 1 (C) 0.5 (D) (s + 2) (s + 2) 2002

6.84

TWO MARKS

The system shown in the figure remains stable when (A) k < - 1 (B) - 1 < k < 3 (C) 1 < k < 3 (D) k > 3

GATE GUIDE and GATE CLOUD by RK Kanodia & Ashish Murolia GATE GUIDEs and CLOUDs are subjectwise books exclusive for GATE EC Examination. Download a sample chapter at

www.nodia.co.in

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

6.85

100 The transfer function of a system is G (s) = . For a (s + 1)( s + 100) unit - step input to the system the approximate settling time for 2% criterion is

(A)100 sec (C) 1 sec 6.86

Page 136

(B) 4 sec (D) 0.01 sec

The characteristic polynomial of a system is 5

4

3

(A) only if 0 # k # 1 (C) only if k > 5

2

q (s) = 2s + s + 4s + 2s + 2s + 1 The system is (A) stable (C) unstable

(B) marginally stable (D) oscillatory

2001 6.92

6.87

The system with the open loop transfer 1 has a gain margin of G (s) H (s) = 2 (A) - 6 db s (s + s + 1) (B) 0 db (C) 35 db (D) 6 db 2001

6.88

6.89

function

ONE MARK

The Nyquist plot for the open-loop transfer function G (s) of a unity negative feedback system is shown in the figure, if G (s) has no pole in the right-half of s -plane, the number of roots of the system characteristic equation in the right-half of s -plane is (A) 0 (B) 1 (C) 2 (D) 3

A I D

Z3 (s) - Z3 (s) , in Z1 (s) + Z3 (s) + Z4 (s) Z1 (s) + Z3 (s) . The equivalent of the block diagram in the figure is given is o - Z (s) - Z (s) c ia. (B) Z2 (s) - Z3 (3s) + Z4 (s) , Z1 (s) +3 Z3 (s) d .no w Z3 (s) Z3 (s) (C) , ww Z2 (s) + Z3 (s) + Z4 (s) Z1 (s) + Z3 (s) - Z3 (s) Z3 (s) (D) , Z2 (s) - Z3 (s) + Z4 (s) Z1 (s) + Z3 (s) 6.93 The open-loop DC gain of a unity negative feedback system with closed-loop transfer function 2 s + 4 is s + 7s + 13 (A) 4 (B) 4 13 9

O N

(A)

6.94

6.91

TWO MARK

An electrical system and its signal-flow graph representations are shown the figure (A) and (B) respectively. The values of G2 and H , respectively are

(C) 4

6.90

(B) only if 1 < k < 5 (D) if 0 # k < 1 or k > 5

If the characteristic equation of a closed - loop system is s2 + 2s + 2 = 0 , then the system is (A) overdamped (B) critically damped (C) underdamped (D) undamped

The feedback control system in the figure is stable

(A) for all K $ 0 (C) only if 0 # K < 1

The root-locus diagram for a closed-loop feedback system is shown in the figure. The system is overdamped.

2000 6.95

(D) 13

(B) only if K $ 0 (D) only if 0 # K # 1 ONE MARK

An amplifier with resistive negative feedback has tow left half plane poles in its open-loop transfer function. The amplifier (A) will always be unstable at high frequency (B) will be stable for all frequency (C) may be unstable, depending on the feedback factor (D) will oscillate at low frequency.

GATE Electronics & Communication by RK Kanodia (Now in 3 Volumes) Purchase Online from our portal www.nodia.co.in and get maximum available discount

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

2000 6.96

TWO MARKS

6.98

ONE MARK

The gain margin (in dB) of a system a having the loop transfer function

(A) 0 (C) 6

2 is s (s + 1) (B) 3 (D) 3

1998

(A) controllable and observable (B) controllable, but not observable (C) observable, but not controllable (D) neither controllable nor observable

A I D

O N

6.102

6.103

TWO MARKS

An amplifier is assumed to have a single-pole high-frequency transfer function. The rise time of its output response to a step function input is 35 n sec . The upper 3 dB frequency (in MHz) for the amplifier to as sinusoidal input is approximately at (A) 4.55 (B) 10 (C) 20 (D) 28.6 If the closed - loop transfer function T (s) of a unity negative feedback system is given by an - 1 s + an T (s) = n n-1 s + a1 s + .... + an - 1 s + an then the steady state error for a unit ramp input is (A) an (B) an an - 1 an - 2 (C) an - 2 (D) zero an - 2 Consider the points s1 =- 3 + j4 and s2 =- 3 - j2 in the s-plane.

.

ia od

6.107

6.108

6.101

The transfer function of a tachometer is of the form (B) K (A) Ks s K (C) (D) n(s K i . + 1) s (s + 1) co

6.106

The phase margin (in degrees) of a system having the loop transferw.n ww function G (s) H (s) = 2 3 is s (s + 1) (A) 45c (B) - 30c (C) 60c (D) 30c 1999

ONE MARK

The number of roots of s3 + 5s2 + 7s + 3 = 0 in the left half of the s -plane is (A) zero (B) one (C) two (D) three

6.105

The system modeled described by the state equations is 0 1 0 x + > Hu X => H 2 -3 1 Y = 81 1B x

6.100

For the system described by the state equation R0V R 0 1 0V W S W S xo = S 0 0 1W x + S0W u SS1WW SS0.5 1 2WW X T X T If the control signal u is given by u = [- 0.5 - 3 - 5] x + v , then the eigen values of the closed-loop system will be (A) 0, - 1, - 2 (B) 0, - 1, - 3 (C) - 1, - 1, - 2 (D) 0, - 1, - 1

6.104

For a second order system with the closed-loop transfer function T (s) = 2 9 s + 4s + 9 the settling time for 2-percent band, in seconds, is (A) 1.5 (B) 2.0 (C) 3.0 (D) 4.0

G (s) H (s) =

6.99

Then, for a system with the open-loop transfer function G (s) H (s) = K 4 (s + 1) (A) s1 is on the root locus, but not s2 (B) s2 is on the root locus, but not s1 (C) both s1 and s2 are on the root locus (D) neither s1 nor s2 is on the root locus

1 A system described by the transfer function H (s) = 3 2 s + as + ks + 3 is stable. The constraints on a and k are. (A) a > 0, ak < 3 (B) a > 0, ak > 3 (C) a < 0, ak > 3 (D) a > 0, ak < 3 1999

6.97

Page 137

6.109

6.110

Consider a unity feedback control system with open-loop transfer K . function G (s) = s (s + 1) The steady state error of the system due to unit step input is (A) zero (B) K (C) 1/K (D) infinite

The transfer function of a zero-order-hold system is (A) (1/s) (1 + e-sT ) (B) (1/s) (1 - e-sT ) (C) 1 - (1/s) e-sT (D) 1 + (1/s) e-sT In the Bode-plot of a unity feedback control system, the value of phase of G (jw) at the gain cross over frequency is - 125c. The phase margin of the system is (A) - 125c (B) - 55c (C) 55c (D) 125c Consider a feedback control system with loop transfer function K (1 + 0.5s) G (s) H (s) = s (1 + s) (1 + 2s) The type of the closed loop system is (A) zero (B) one

GATE GUIDE and GATE CLOUD by RK Kanodia & Ashish Murolia GATE GUIDEs and CLOUDs are subjectwise books exclusive for GATE EC Examination. Download a sample chapter at

www.nodia.co.in

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

Page 138

(C) two (D) three 6.111

6.112

6.113

6.114

The transfer function of a phase lead controller is 1 + 3Ts . The 1 + Ts maximum value of phase provided by this controller is (A) 90c (B) 60c (C) 45c (D) 30c The Nyquist plot of a phase transfer function g (jw) H (jw) of a system encloses the (–1, 0) point. The gain margin of the system is (A) less than zero (B) zero (C) greater than zero (D) infinity 2s2 + 6s + 5 (s + 1) 2 (s + 2) The characteristic equation of the system is (A) 2s2 + 6s + 5 = 0 (B) (s + 1) 2 (s + 2) = 0 (C) 2s2 + 6s + 5 + (s + 1) 2 (s + 2) = 0 (D) 2s2 + 6s + 5 - (s + 1) 2 (s + 2) = 0 The transfer function of a system is

A I D

In a synchro error detector, the output voltage is proportional to [w (t)] n, where w (t) is the rotor velocity and n equals (A) –2 (B) –1 (C) 1 (D) 2

O N

no w.

d

w

w ONE MARK

1997 6.115

.in

co ia.

In the signal flow graph of the figure is y/x equals

(A) 3 (B) 5 2 (C) 2 (D) None of the above 6.116

A certain linear time invariant system has the state and the output equations given below 1 - 1 X1 0 Xo1 > o H = >0 1 H>X H + >1H u 2 X2 y = 81 1B: X1 D X2 If X1 (0) = 1, X2 (0) =- 1, u (0) = 0, then

dy dt

is t=0

(A) 1 (B) –1 (C) 0 (D) None of the above ***********

GATE Electronics & Communication by RK Kanodia (Now in 3 Volumes) Purchase Online from our portal www.nodia.co.in and get maximum available discount

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

Page 139

SOLUTIONS 6.1

Pk2 = ^1 h^s-1h^1 h^1 h = s-1 since, all the loops are touching to the paths Pk1 and Pk2 so, Dk 1 = Dk 2 = 1 Now, we have D = 1 - (sum of individual loops) + (sum of product of nontouching loops) Here, the loops are

Option (B) is correct. From the given plot, we obtain the slope as 20 log G2 - 20 log G1 Slope = log w2 - log w1

L1 = ^- 4h^1 h =- 4 L2 = ^- 4h^s-1h = 4s-1 L 3 = ^- 2h^s-1h^s-1h =- 2s-2 L 4 = ^- 2h^s-1h^1 h =- 2s-1 As all the loop L1, L2, L 3 and L 4 are touching to each other so, D = 1 - ^L1 + L2 + L 3 + L 4h

From the figure

and

20 log G2 20 log G1 w1 w2

=- 8 dB = 32 dB = 1 rad/s = 10 rad/s

So, the slope is

Slope = - 8 - 32 log 10 - log 1 =- 40 dB/decade Therefore, the transfer function can be given as G ^s h = k2 S at w = 1 G ^ jwh = k 2 = k w In decibel, 20 log G ^ jwh = 20 log k = 32 32

6.2

or, k = 10 = 39.8 Hence, the Transfer function is G ^s h = k2 = 392.8 s s Option (C) is correct. Given, open loop transfer function G ^s h = 10Ka = Ka 1 1 + 10s s + 10 By taking inverse Laplace transform, we have g ^ t h = e- t

= 1 - ^- 4 - 4s-1 - 2s-2 - 2s-1h = 5 + 6s1 + 2s2 From Mason’s gain formulae Y ^s h = SPk Dk D U ^s h s-2 + s-1 = 5 + 6s-1 + 2s-2 = 2s+1 5s + 6s + 2

A I D

20

1 10

O N

.in

no w.

ww

6.3

Option (A) is correct. For the given SFG, we have two forward paths

d

and

y = ^- 1h^1 h x2 + ^- 1h^1 h^- 1h x1 + ^1 h^- 1h^1 h^- 1h^1 h u = x1 - x 2 + u Hence, in matrix form we can write the state variable equations - 1 0 x1 -1 xo1 > o H = > 1 - 1H >x H + > 1 H u x2 2 x1 and y = 81 - 1B > H + u x2

1 10

Now, given that ka reduces open loop time constant by a factor of 100. i.e., tcl = tol 100 1 or, = 10 100 ka Hence, ka = 10

co ia.

So, from the state diagram, we obtain xo1 =- x1 - u xo2 =- x2 + ^1 h^- 1h^1 h^- 1h u + ^- 1h^1 h^- 1h x1 xo2 =- x2 + x1 + u

Comparing with standard form of transfer function, Ae-t/t , we get the open loop time constant, tol = 10 Now, we obtain the closed loop transfer function for the given system as G ^s h 10Ka H ^s h = = 1 + G ^s h 1 + 10s + 10Ka Ka = s + ^Ka + 101 h By taking inverse Laplace transform, we get h ^ t h = ka .e-^k + ht So, the time constant of closed loop system is obtained as tcl = 1 1 ka + 10 or, tcl = 1 ka (approximately) a

Option (A) is correct. For the shown state diagram we can denote the states x1 , x2 as below

6.4

which can be written in more general form as -1 0 -1 X +> H Xo = > 1 - 1H 1 y = 81 - 1B X + u 6.5

Option (A) is correct. From the obtained state-variable equations We have -1 0 A => 1 - 1H So,

S+1 0 SI - A = > - 1 S + 1H

Pk1 = ^1 h^s-1h^s-1h^1 h = s-2

GATE GUIDE and GATE CLOUD by RK Kanodia & Ashish Murolia GATE GUIDEs and CLOUDs are subjectwise books exclusive for GATE EC Examination. Download a sample chapter at

www.nodia.co.in

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

1 >S + 1 0 H ^S + 1h2 1 S + 1 R 1 V S 0 W S+1 W =S 1 1 W S S^S + 1h2 S + 1W T X Hence, the state transition matrix is obtained as eAt = L-1 ^SI - Ah-1 V_ ZR 1 ]]S 0 Wbb S+1 W = L-1 [S 1 1 W` S ]S^S + 1h2 S + 1Wb \T Xa e-1 0 = > -t -tH te e Option (C) is correct. (s2 + 9) (s + 2) G (s) = (s + 1) (s + 3) (s + 4) (- w2 + 9) (jw + 2) = (jw + 1) (jw + 3) (jw + 4) The steady state output will be zero if

^SI - Ah-1 =

and

6.6

G (jw) = 0 -w 2 + 9 = 0 6.7

Page 140

&

R 0 a 0VR0V R 0V 1 WS W S W S AB = S 0 0 a2WS0W = Sa2W SSa 0 0WWSS1WW SS 0WW 3 RT 0 0XT aX1 a2VWTRS0XVW RSa1 a2VW S A2 B = Sa2 a 3 0 0WS0W = S 0W SS 0 a a 0WWSS1WW SS 0WW 3 1 T XT X T X For controllability it is necessary that following matrix has a tank of n = 3 . R0 0 a a V 1 2W S 2 U = 6B : AB : A B@ = S0 a2 0W SS1 0 0WW So, a2 ! 0 X T a1 a 2 ! 0 & a1 ! 0 a 3 may be zero or not. 6.9

w = 3 rad/s

6.10

Option (A) is correct.

Option (B) is correct. For given plot root locus exists from - 3 to 3, So there must be odd number of poles and zeros. There is a double pole at s =- 3 Now poles = 0, - 2, - 3, - 3 zeros =- 1 k (s + 1) Thus transfer function G (s) H (s) = s (s + 2) (s + 3) 2 Option (A) is correct. We have G (jw) = 5 + jw Here s = 5 . Thus G (jw) is a straight line parallel to jw axis.

K (s + 1) [R (s) - Y (s)] s3 + as2 + 2s + 1 6.11 Option (B) is correct. K (s + 1) K (s + 1) R (s) Y (s) ;1 + 3 = Here x s + as2 + 2s + 1E s3 + as2 + 2s + 1 3 2 Y (s) [s + as + s (2 + k) + (1 + k)] = K (s + 1) R (s) y Y (s) Transfer Function, H (s) = R (s) .in oNow K (s + 1) c . y1 = 3 ia d s + as2 + s (2 + k) + (1 + k) .no y1 (s + 2) Routh Table : w ww yo1 + 2y1 xo + 2x xo xo Y (s) =

A I D

O N

For oscillation,

Auxiliary equation

=u =u =u =- 2x + u = [- 2] x + [1] u Drawing SFG as shown below

a (2 + K) - (1 + K) =0 a a = K+1 K+2 A (s) = as2 + (k + 1) = 0 s2 =- k + 1 = - k + 1 (k + 2) =- (k + 2) a (k + 1)

xo1 = [- 2] x1 + [1] u y1 = x1 ; y2 = 2x1

Thus

s = j k+2

y1 1 y = > H = > H x1 y2 2

jw = j k + 2

and 6.8

w = k + 2 = 2 (Oscillation frequency) k =2 a = 2 + 1 = 3 = 0.75 2+2 4

Option (D) is correct. General form of state equations are given as xo = Ax + Bu yo = Cx + Du For the given problem R 0 a 0V R0V 1 S W S W A = S 0 0 a2W, B = S0W SSa SS1WW 0 0WW 3 T T X X & Communication GATE Electronics

dy = y1 and xo = 1 dx 1 y1 x = > H = > H = > Hx 2 y2 2x = 1 u s+2

x1 = x

Here 6.12

Option (C) is correct. 100 s (s + 10) 2 100 Now G (jw) H (jw) = jw (jw + 10) 2 If wp is phase cross over frequency +G (jw) H (jw) = 180c We have

G (s) H (s) =

Thus

- 180c = 100 tan-1 0 - tan-1 3 - 2 tan-1 a

or or

- 180c =- 90 - 2 tan-1 (0.1wp) 45c = tan-1 (0.1wp)

by RK Kanodia (Now in 3 Volumes) Purchase Online from our portal www.nodia.co.in and get maximum available discount

wp 10 k

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

or or Now

Page 141

tan 45c 0.1wp = 1 wp = 10 rad/se 100 G (jw) H (jw) = w (w2 + 100)

At w = wp

So,

100 G (jw) H (jw) = = 1 10 (100 + 100) 20 Gain Margin =- 20 log 10 G (jw) H (jw) =- 20 log 10 b 1 l 20 = 26 dB 6.13

6.14

Option (D) is correct. From option (D) TF = H (s) 100 100 = ! s (s2 + 100) s (s + 10) 2

Option (A) is correct. Initial slope is zero, so K = 1 At corner frequency w 1 = 0.5 rad/ sec , slope increases by + 20 dB/ decade, so there is a zero in the transfer function at w 1 At corner frequency w 2 = 10 rad/ sec , slope decreases by - 20 dB/ decade and becomes zero, so there is a pole in transfer function at w2 K a1 + s k w1 Transfer function H (s) = s a1 + w 2 k 1 a1 + s k (1 + 10s) 0.1 = = s (1 + 0.1s) 1 + a 0. 1 k 6.17 Option (D) is correct. Steady state error is given as sR (s) eSS = lim n s " 0 i 1 + G (s) GC (s) co. . a (unit step unit) R (s) = 1 di o s n . ...(1)w 1 ww eSS = lim s " 0 1 + G (s) GC (s) ...(2) 1 = lim s"0 GC (s) 1+ 2 s + 2s + 2 eSS will be minimum if lim GC (s) is maximum s"0 In option (D) lim GC (s) = lim 1 + 2 + 3s = 3 s s"0 s"0 So, eSS = lim 1 = 0 (minimum) s"0 3

A I D

E (s) = R (s) - H (s) E (s) (s + 1)

1 = R (s) - Y (s) s + 1D sE (s) = R (s) - Y (s) (s + 1) E (s) Y (s) = s+1

E (s) :1 -

O N

sY (s) = R (s) - Y (s) (s + 1) Y (s) = R (s)

From (1) and (2) Transfer function

Y (s) = 1 R (s) s + 1 6.15

p = 2/ 3

6.16

1 s+1

= R (s) - Y (s) +

qh = 9- p - a- p kC = p 3 2 6 p = p - tan-1 w apk 2 6 tan-1 a w k = p - p = p p 2 6 3 w = tan p = 3 a3k p 2 = 3 , (w = 2 rad/ sec) p

or

Option (B) is correct. From the given block diagram

H (s) = Y (s) - E (s) $

p = 2/ 3

or Alternative :

Option (B) is correct. Transfer function is given as Y (s) = s X (s) s + p jw H (jw) = jw + p

6.18

H (s) =

Option (D) is correct. Assign output of each integrator by a state variable

Amplitude Response H (jw) = Phase Response Input Output

or

w w2+ p 2

qh (w) = 90c - tan-1 a w k p x (t) = p cos a2t - p k 2 y (t) = H (jw) x (t - qh) = cos a2t - p k 3 w H (jw) = p = 2 w +p2 1 = 2 , (w = 2 rad/ sec) p 4+p2 4p 2 = 4 + p 2 & 3p 2 = 4

xo1 =- x1 + x2 xo2 =- x1 + 2u y = 0.5x1 + 0.5x2 State variable representation -1 1 0 x + > Hu xo = > H -1 0 2 yo = [0.5 0.5] x 6.19

Option (C) is correct.

GATE GUIDE and GATE CLOUD by RK Kanodia & Ashish Murolia GATE GUIDEs and CLOUDs are subjectwise books exclusive for GATE EC Examination. Download a sample chapter at

www.nodia.co.in

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

Page 142

By masson’s gain formula

Transfer function Y (s) H (s) = = U (s) Forward path given

- qd - qp1 + qz1 + qz2 = 180c - qd = 180c - (- qp1 + qz1 + q2) = 180c - (90c + 180 - 45c) =- 45c

/ PK DK D

6.23

P1 (abcdef ) = 2 # 1 # 1 # 0.5 = 12 s s s P2 (abcdef ) = 2 # 1 # 1 # 0.5 3 Loop gain L1 (cdc) =- 1 s L2 (bcdb) = 1 # 1 # - 1 = -21 s s s

kwn2 where x < 1 s + 2xwn s + wn2 Thus (A) or (B) may be correct For option (A) wn = 1.12 and 2xwn = 2.59 " x = 1.12 For option (B) wn = 1.91 and 2xwn = 1.51 " x = 0.69 T (s) =

6.24

D = 1 - [L1 + L2] = 1 - :- 1 - 12 D = 1 + 1 + 12 s s s s So,

6.20

6.21

D1 = 1, D2 = 2 Y (s) H (s) = = P1 D 1 + P2 D 2 D U (s) 1 :1+1:1 2 (1 + s) s =s = 2 1 1 (s + s + 1) 1+ + 2 s s

6.25

2

Option (B) is correct. The plot has one encirclement of origin in clockwise direction. Thus G (s) has a zero is in RHP. Option (C) is correct. The Nyzuist plot intersect the real axis ate - 0.5. Thus G. M. =- 20 log x =- 20 log 0.5 = 6.020 dB And its phase margin is 90c.

A I D

Option (C) is correct. Option (C) is correct. Transfer function for the given pole zero plot is: (s + Z1)( s + Z2) This compensator is roughly equivalent to combining lead and lad (s + P1)( s + P2) compensators in the same design and it is referred also as PID compensator. .in the plot Re (P1 and P2 )>(Z1 and Z2 ) oFrom c . Option (C) is correct. dia So, these are two lead compensator. o Hence both high pass filters and the system is high pass filter. .n p 1 0 w Here and B = = G A == ww 6.27 Option (C) is correct. q 0 1G 1 0 p p == G G G = 0 1 q q p q S = 8B AB B = = q pG

AB = =

6.26

O N

Percent overshoot depends only on damping ratio, x . 2

Mp = e- xp 1 - x If Mp is same then x is also same and we get x = cos q Thus q = constant The option (C) only have same angle.

S = pq - pq = 0 Since S is singular, system is completely uncontrollable for all values of p and q . 6.22

Option (B) is correct. For under-damped second order response

Option (B) is correct. The characteristic equation is 1 + G (s) H (s) = 0 K (s2 - 2s + 2) or =0 1+ s2 + 2s + 2 or s2 + 2s + 2 + K (s2 - 2s + 2) = 0 2 or K =- s2 + 2s + 2 s - 2s + 2 For break away & break in point differentiating above w.r.t. s we have 2 2 dK =- (s - 2s + 2)( 2s + 2) - (s + 2s + 2)( 2s - 2) = 0 ds (s2 - 2s + 2) 2

Thus

6.28

6.29

Option (D) is correct. P = 2 25 2xwn = 0, x = 0 " Undamped s + 25

Graph 3

Q=

62 s + 20s + 62

2xwn = 20, x > 1 " Overdamped

Graph 4

R=

62 s2 + 12s + 62

2xwn = 12, x = 1 " Critically

Graph 1

S=

72 s2 + 7s + 72

2xwn = 7, x < 1 " underdamped

Graph 2

2

Option (C) is correct. We labeled the given SFG as below :

(s2 - 2s + 2)( 2s + 2) - (s2 + 2s + 2)( 2s - 2) = 0

or s =! 2 Let qd be the angle of departure at pole P , then From this SFG we have

GATE Electronics & Communication by RK Kanodia (Now in 3 Volumes) Purchase Online from our portal www.nodia.co.in and get maximum available discount

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

6.30

xo1 =- gx1 + bx3 + m1 xo2 = gx1 + ax3 xo3 =- bx1 - ax3 + u2 R V R VR V R V Sx1 W S- g 0 b WSx1 W S0 1 W u1 Sx2 W = S g 0 a WSx2 W+ S0 0 We o Thus SSx WW SS- b 0 - a WWSSx WW SS1 0 WW u2 3 3 X T X T XT X T Option (C) is correct. The characteristic equation of closed lop transfer function is

Page 143

For R,

Since R2 C2 > R1 C1, it is lag compensator. Option (D) is correct. In a minimum phase system, all the poles as well as zeros are on the left half of the s -plane. In given system as there is right half zero (s = 5), the system is a non-minimum phase system.

6.34

1 + G (s) H (s) = 0 =0 1+ 2 s+8 s + as - 4 or s 2 + as - 4 + s + 8 = 0 or s2 + (a + 1) s + 4 = 0 This will be stable if (a + 1) > 0 " a > - 1. Thus system is stable for all positive value of a. 6.31

Option (B) is correct. We have Kv = lim sG (s) H (s)

6.35

s"0

1 + G (s) H (s) = 0 1000 = lims " 0 s

or

3

6

2

z4

5

3

1

3

21 5

7 5

z2

4 3

3

z1

- 74

z0

1

N

w

.no

ww

d

1

For Q ,

Z =

(sC2 R2 + 1) sC2

We have

5 (s + 5)( s2 + s + 1) 5 = = 2 1 2 s 5`1 + j (s + s + 1) s +s+1 5

T (s) =

Option (A) is correct. 1 = 1 ( s 1 )( s - 1) + s -1 The lead compensator C (s) should first stabilize the plant i.e. 1 remove term. From only options (A), C (s) can remove this (s - 1) term

5 2x 1 - x2

1

.in (D) is correct. Option

G (s) =

wr = wn 1 - 2x2

Option (B) is correct. The given circuit is a inverting amplifier and transfer function is Vo = - Z = - Z (sC1 R1 + 1) R Vi R1 sC R + 1 1

or

2xwn = 10 + 100KD KD = 0.9

. We can see easily that pole at s =- 0.5 ! j 23 is dominant then pole at s =- 5 . Thus we have approximated it. 6.37

We have wr = 5 2 , and mr = 10 . Only options (A) satisfy these 3 values. wn = 10, x = 1 2 where wr = 10 1 - 2` 1 j = 5 2 4 and Hence satisfied mr = 1 5 1 = 10 22 1- 4 3

Kp = 100

In given transfer function denominator is (s + 5)[( s + 0.5) 2 + 43 ]

Option (C) is correct. For underdamped second order system the transfer function is Kwn2 s2 + 2xwn s + wn2 It peaks at resonant frequency. Therefore

1 + G (s) H (s) = 0 (100 + KD s) 100 =0 1+ s (s + 10)

or s2 + (10 + 100KD) s + 10 4 = 0 Comparing with s2 + 2xwn + wn2 = 0 we get

co ia.

6.36

T (s) =

6.33

IA

D O

z5

mr =

(Kp + KD s) 100 = Kp s (s + 100)

Now characteristics equation is

3z5 + 5z 4 + 6z3 + 3z2 + 2z + 1 = 0 The routh table is shown below. As there are tow sign change in first column, there are two RHS poles.

Resonant frequency and peak at this frequency

s"0

(Kp + KD s) 100 = Kp s (s + 100)

Now characteristics equations is

1 + G (s) = 0 5 4 or s + 2s + 3s3 + 6s2 + 5s + 3 = 0 Substituting s = z1 we have

6.32

1000 = lim s

or

Option (C) is correct. The characteristic equation is

z

Vo =- (sC2 R2 + 1) (sC1 R1 + 1) PID Controller # Vi sC2 R1 R2 Z = (sC2 R2 + 1) (sC1 R1 + 1) Vo =R2 # Vi (sC2 R2 + 1) R1

Thus

satisfies. 6.38

2

10 (s - 1) 1 # (s + 1)( s - 1) (s + 2) 10 Only option (A) = (s + 1)( s + 2)

G (s) C (s) =

Option (D) is correct. For ufb system the characteristics equation is 1 + G (s) = 0 K or 1+ =0 2 s (s + 7s + 12) or s (s2 + 7s + 12) + K = 0 Point s =- 1 + j lie on root locus if it satisfy above equation i.e (- 1 + j)[( - 1 + j) 2 + 7 (- 1 + j) + 12) + K] = 0 or K =+ 10

GATE GUIDE and GATE CLOUD by RK Kanodia & Ashish Murolia GATE GUIDEs and CLOUDs are subjectwise books exclusive for GATE EC Examination. Download a sample chapter at

www.nodia.co.in

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia 6.39

Page 144

Solving (1) and (2) set of equations we get p q 0 1 =r s G = =- 2 - 3G

Option (D) is correct. At every corner frequency there is change of -20 db/decade in slope which indicate pole at every corner frequency. Thus K G (s) = s (1 + s)`1 + s j 20

The characteristic equation lI - A = 0

Bode plot is in (1 + sT) form 20 log K = 60 dB = 1000 w w = 0. 1

6.40

or l (l + 3) + 2 = 0 or l =- 1, - 2 Thus Eigen values are - 1 and - 2 Eigen vectors for l1 =- 1

K =5

Thus Hence

l -1 =0 2 l+3

G (s) =

100 s (s + 1)( 1 + .05s)

Option (A) is correct. We have or and

dw dt dia dt

> H = =- 1 - 10G=in G + =10Gu -1

1 w

0

(l1 I - A) X1 = 0

dw =- w + i n dt dia =- w - 10i + 10u a dt

...(2)

or - x11 - x21 = 0 or x11 + x21 = 0 We have only one independent equation x11 =- x21. Let x11 = K , then x21 =- K , the Eigen vector will be x11 K 1 =x G = =- K G = K =- 1G 21

Taking Laplace transform (i) we get sw (s) =- w (s) = Ia (s) or (s + 1) w (s) = Ia (s) Taking Laplace transform (ii) we get or or or or 6.41

...(3)

sIa (s) =- w (s) - 10Ia (s) + 10U (s) w (s) = (- 10 - s) Ia (s) + 10U (s) = (- 10 - s)( s + 1) w (s) + 10U (s) w (s) =- [s2 + 11s + 10] w (s) + 10U (s) (s2 + 11s + 11) w (s) = 10U (s) w (s) = 2 10 U (s) (s + 11s + 11)

Thus

e-2t > d (- 2e-2t)H dt

t=0

or

p - 2q =- 2 and r - 2s = 4 1 For initial state vector x (0) = = G the system response is -1 e-t x (t) = > -tH -e

Thus

6.42

p q 1 == r s G=- 2G

We get

e-t > d (- e-t)H dt

t=0

- e- (0)

==

p q

6.43

6.44

...(i)

6.45

or - x11 - x21 = 0 or x11 + x21 = 0 We have only one independent equation x11 =- x21. Let x11 = K, then x21 =- K , the Eigen vector will be x12 K 1 =x G = =- 2K G = K =- 2G 22 Option (D) is correct. As shown in previous solution the system matrix is 0 1 A == - 2 - 3G Option (D) is correct. Given system is 2nd order and for 2nd order system G.M. is infinite. Option (D) is correct. Option (D) is correct. If the Nyquist polt of G (jw) H (jw) for a closed loop system pass through (- 1, j0) point, the gain margin is 1 and in dB GM =- 20 log 1 = 0 dB

p q 1 r s G=- 1G

6.46

Option (B) is correct. The characteristics equation is 1 + G (s) H (s) = 0

1

> e- (0) H = =r s G=- 1G

1+

-1 p-q = 1G = = r - s G

We get

Now Eigen vector for l2 =- 2 (l2 I - A) X2 = 0 l2 - 1 x12 or = 2 l + 3G=x G = 0 2 22 n i - 2 - 1 x11 coor. =0 . = a 2 1 G=x21G di

w

p q 1 - 2e-2 (0) > 4e-2 (0) H = =r s G=- 2G -2 p - 2q = 4 G = = r - 2s G

d dt

no

. ww

1 For initial state vector x (0) = = G the system response is -2 e-2t x (t) = > H - 2e-2t d dt

A I D

From (3)

O N

Option (A) is correct. We have xo (t) = Ax (t) p q Let A == r sG

l1 - 1 x11 = 2 l + 3G=x G = 0 1 21 - 1 - 1 x11 = 2 2 G=x G = 0 21

or

...(1)

p - q =- 1 and r - s = 1

...(2)

K (s + 1) =0 s + as2 + 2s + 1 3

s3 + as2 + (2 + K) s + K + 1 = 0

GATE Electronics & Communication by RK Kanodia (Now in 3 Volumes) Purchase Online from our portal www.nodia.co.in and get maximum available discount

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

The Routh Table is shown below. For system to be oscillatory stable a (2 + K) - (K + 1) =0 a or ...(1) a = K+1 K+2

Page 145

Option (C) is correct. For a = 0.84 we have

6.50

G (s) = 0.84s2 + 1 s Due to ufb system H (s) = 1 and due to unit impulse response R (s) = 1, thus C (s) = G (s) R (s) = G (s) = 0.84s2 + 1 = 12 + 0.84 s s s Taking inverse Laplace transform

Then we have as2 + K + 1 = 0 At 2 rad/sec we have s = jw " s2 =- w2 =- 4 , Thus - 4a + K + 1 = 0 Solving (i) and (ii) we get K = 2 and a = 0.75 .

6.47

...(2) At t = 1,

s3

1

2+K

s2

a

1+K

s1

(1 + K) a - (1 + K) a

s0

1+K

6.48

N

Option (A) is correct. s 0 0 1 s -1 (sI - A) = = -= == G G 0 s -1 0 1 sG (sI - A)

s s -1 s +1 1 = 2 G = > -1 = s +1 1 s s +1 2

2

1 s2 + 1 s s2 + 1

or

6.53

6.54 6.55

Thus or or

1+a wg2

6.56

wg = (2)

Option (A) is correct. Despite the presence of negative feedback, control systems still have problems of instability because components used have nonlinearity. There are always some variation as compared to ideal characteristics. Option (B) is correct. Option (C) is correct. The peak percent overshoot is determined for LTI second order closed loop system with zero initial condition. It’s transfer function is 2

Option (A) is correct.

s"0

R (s) 1 = lim 1 + G (s) s " 0 s + sG (s) ess = lim 1 = 5% = 1 s " 0 sG (s) 20 kv = 1 = lim sG (s) = 20 s"0 ess = lim s s"0

or But

Finite

kv is finite for type 1 system having ramp input.

(as awg = 1) 1 4

+T (jw) = 0

For ramp input we have R (s) = 12 s Now ess = lim sE (s)

=1

1 + 1 = wg2

At w = 3 ,

+T (jw) = tan-1 (wT) - tan-1 (wbT) T (jw) = 1 +T (jw) =- tan-1 0 = 0 T (jw) = 1 b

wn2 s + 2xwn s + wn2 Transfer function has a pair of complex conjugate poles and zeroes.

or awg = 1 At gain crossover frequency G (jwg) = 1 wg2

At w = 3 ,

b > 1; T > 0

1 + w2 T2 1 + w2 b2 T2

T (jw) =

T (s) =

p = p + tan-1 (w a) - p g 4 p = tan-1 (w a) g 4

2

and At w in = 0 , . o a.cAt w = 0 ,

di

H

Option (C) is correct. 1 G (s) = as + We have s2 +G (jw) = tan-1 (wa) - p Since PM is p i.e. 45c, thus 4 p = p + +G (jw ) w " Gain cross over Frequeng g 4 cy or

.no

w ww

cos t sin t f (t) = eAt = L-1 [(sI - A)] -1 = = - sin t cos t G 6.49

IA

D O

fmax = tan-1 a - 1 2 a -1 3 - 1 = tan = tan-1 1 2 3 3 fmax = p 6

-1

Option (D) is correct. The transfer function of a lag network is T (s) = 1 + sT 1 + sbT

6.52

T>0

The maximum phase sift is

or

Option (C) is correct. We have where l is set of Eigen values Xo = AX + BU o and where m is set of Eigen values W = CW + DU If a liner system is equivalently represented by two sets of state equations, then for both sets, states will be same but their sets of Eigne values will not be same i.e. X = W but l ! m

6.51

Option (D) is correct. The transfer function of given compensator is Gc (s) = 1 + 3Ts 1 + Ts Comparing with Gc (s) = 1 + aTs we get a = 3 1 + Ts

c (t) = (t + 0.84) u (t) c (1 sec) = 1 + 0.84 = 1.84

6.57

Option (A) is correct.

GATE GUIDE and GATE CLOUD by RK Kanodia & Ashish Murolia GATE GUIDEs and CLOUDs are subjectwise books exclusive for GATE EC Examination. Download a sample chapter at

www.nodia.co.in

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia 6.58

Option (C) is correct. Any point on real axis of s - is part of root locus if number of OL poles and zeros to right of that point is even. Thus (B) and (C) are possible option. The characteristics equation is

Page 146

or 6.61

1 + G (s) H (s) = 0 or or

1+

K (1 - s) =0 s (s + 3) 2 K = s + 3s 1-s

For break away & break in point dK = (1 - s)( 2s + 3) + s2 + 3s = 0 ds or - s2 + 2s + 3 = 0 which gives s = 3 , - 1 Here - 1 must be the break away point and 3 must be the break in point. 6.59

Option (D) is correct. -2s G (s) = 3e s (s + 2) -2jw or G (jw) = 3e jw (jw + 2) 3 G (jw) = w w2 + 4 Let at frequency wg the gain is 1. Thus 3 =1 wg (wg2 + 4)

or or or Now

wg4 + 4wg2 - 9 = 0 wg2 = 1.606 wg = 1.26 rad/sec +G (jw) =- 2w - p - tan-1 w 2 2

Let at frequency wf we have +GH =- 180c w - p =- 2wf - p - tan-1 f 2 2 w or 2wf + tan-1 f = p 2 2 w w 3 or 2wf + c f - 1 ` f j m = p 2 2 3 2 or

or 6.60

6.62

Option (D) is correct. The open loop transfer function is 2 (1 + s) G (s) H (s) = s2 Substituting s = jw we have 2 (1 + jw) ...(1) G (jw) H (jw) = - w2 +G (jw) H (jw) =- 180c + tan-1 w The frequency at which phase becomes - 180c, is called phase crossover frequency. Thus - 180 =- 180c + tan-1 wf or tan-1 wf = 0 or wf = 0 The gain at wf = 0 is w2 = 3 G (jw) H (jw) = 2 1 + 2 w 1 Thus gain margin is = = 0 and in dB this is - 3 . 3 Option (C) is correct. Centroid is the point where all asymptotes intersects. SReal of Open Loop Pole - SReal Part of Open Loop Pole s = SNo.of Open Loop Pole - SNo.of Open Loop zero = - 1 - 3 =- 1.33 3

A I D 6.63

O N

Option (C) is correct. The given bode plot is shown below

.in

no w.

co ia.

d

ww

At w = 1 change in slope is +20 dB " 1 zero at w = 1 At w = 10 change in slope is - 20 dB " 1 poles at w = 10 At w = 100 change in slope is - 20 dB " 1 poles at w = 100 K (s + 1) Thus T (s) = s s + 1) ( 10 + 1)( 100 Now 20 log10 K =- 20 " K = 0.1 0.1 (s + 1) 100 (s + 1) Thus = T (s) = s s ( 10 + 1)( 100 + 1) (s + 10)( s + 100)

5wf wf3 =p 2 2 24 5wf .p 2 2

6.64

wf = 0.63 rad

Option (D) is correct. The gain at phase crossover frequency wf is 3 3 G (jwg) = = 2 wf (wf + 4) 0.63 (0.632 + 4) or G (jwg) = 2.27 G.M. =- 20 log G (jwg) - 20 log 2.26 =- 7.08 dB Since G.M. is negative system is unstable. The phase at gain cross over frequency is w +G (jwg) =- 2wg - p - tan-1 g 2 2 =- 2 # 1.26 - p - tan-1 1.26 2 2

=- 4.65 rad or - 266.5c PM = 180c + +G (jwg) = 180c - 266.5c =- 86.5c

1 2

Option (C) is correct. We have r (t) = 10u (t) or R (s) = 10 s Now H (s) = 1 s+2 C (s) = H (s) $ R (s) = or

1 $ 10 10 s + 2 s s (s + 2)

C (s) = 5 - 5 s s+2

c (t) = 5 [1 - e-2t] The steady state value of c (t) is 5. It will reach 99% of steady state value reaches at t , where 5 [1 - e-2t] = 0.99 # 5 or

1 - e-2t = 0.99

GATE Electronics & Communication by RK Kanodia (Now in 3 Volumes) Purchase Online from our portal www.nodia.co.in and get maximum available discount

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

or or 6.65

6.66

Page 147

e-2t = 0.1 - 2t = ln 0.1 t = 2.3 sec

We have only one independent equation x12 = 2x22 Let x22 = K , then x12 = 2K . Thus Eigen vector will be 2 x12 2K =x G = = K G = K = 1 G 22

Option (A) is correct. Approximate (comparable to 90c) phase shift are Due to pole at 0.01 Hz " - 90c Due to pole at 80 Hz " - 90c Due to pole at 80 Hz " 0 Due to zero at 5 Hz " 90c Due to zero at 100 Hz " 0 Due to zero at 200 Hz " 0 Thus approximate total - 90c phase shift is provided.

Digonalizing matrix -1 2 x11 x12 M == = x21 x22 G = 1 1G 1 -2 M-1 = ` - 1 j= G 3 -1 -1

Now

Now Diagonal matrix of sin At is D where sin (l1 t) 0 sin (- 4t) 0 D == == G 0 sin (l2 t) 0 sin (l2 t)G

Option (C) is correct. Mason Gain Formula T (s) =

1 - 2 x12 =- 1 2 G=x G = 0 22

or

Spk 3 k 3

Now matrix

In given SFG there is only one forward path and 3 possible loop.

- 1 2 sin (- 4t) 0 1 -2 =-` 1 j= G G = = 1 1 0 sin (- t) - 1 - 1G 3

p1 = abcd 31 = 1 3= 1 - (sum of indivudual loops) - (Sum of two non touching loops)

- sin (- 4t) - 2 sin (- t) 2 sin (- 4t) - 2 sin (- t) =-` 1 j= sin (- 4t) + 2 sin (t) - 2 sin (- 4t) - sin (- t)G 3

= 1 - (L1 + L2 + L3) + (L1 L3) Non touching loop are L1 and L3 where L1 L2 = bedg Thus

6.67

C (s) p1 3 1 = 1 - (be + cf + dg) + bedg R (s) abcd = 1 - (be + cf + dg) + bedg

Option (A) is correct. We have

-2 2 A == 1 - 3G

Characteristic equation is [lI - A] = 0 or

- sin (- 4t) - 2 sin (- t) 2 sin (- 4t) - 2 sin (- t) =-` 1 j= sin (- 4t) - sin (- t) - 2 sin (- 4t) + 2 sin (- t)G 3

A I D

O N

sin (- 4t) + 2 sin (- t) - 2 sin (- 4t) + 2 sin (- t) = ` 1 j= Gs 3 - sin (- 4t + sin (- t) 2 sin (- 4t) + sin (- t)

no w.

d

ww

or - 2x11 - 2x21 = 0 or x11 + x21 = 0 We have only one independent equation x11 =- x21. Let x21 = K , then x11 =- K , the Eigen vector will be x11 -K -1 =x G = = K G = K = 1 G 21

1 + G (s) = 0 1 + G (s)

K =0 s (s2 + 2s + 2)( s + 3) s 4 + 4s3 + 5s2 + 6s + K = 0 The routh table is shown below. For system to be stable, (21 - 4K) 0 < K and 0 < 2/7 This gives 0 < K < 21 4 1+

l + 2 -2 =0 -1 l + 3

(l1 I - A) X1 = 0 l1 + 2 - 2 x11 = 1 l + 3G=x G = 0 1 21 - 2 - 2 x11 =- 1 - 1G=x G = 0 21

n Option o.i (A) is correct. c . ia For ufb system the characteristic equation is

6.68

or (l + 2)( l + 3) - 2 = 0 or l2 + 5l + 4 = 0 Thus l1 =- 4 and l2 =- 1 Eigen values are - 4 and - 1. Eigen vectors for l1 =- 4 or

B = sin At = MDM-1

6.69

s4

1

5

K

s3

4

6

0

s2

7 2

K

s1

21 - 4K 7/2

0

s0

K

Option (B) is correct. We have P (s) = s5 + s 4 + 2s3 + 3s + 15 The routh table is shown below. If e " 0+ then 2e +e 12 is positive and -15e2-e +2412e - 144 is negative. Thus there are two sign change in first column. Hence system has 2 root on RHS of plane. 2

Now Eigen vector for l2 =- 1 (l2 I - A) X2 = 0 l2 + 2 - 2 x12 or = - 1 l + 3G=x G = 0 2 22

s5

1

2

3

s4

1

2

15

s3

e

- 12

0

GATE GUIDE and GATE CLOUD by RK Kanodia & Ashish Murolia

GATE GUIDEs and CLOUDs are subjectwise books exclusive for GATE EC Examination. Download a sample chapter at

www.nodia.co.in

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

s2

2e + 12 e

s1

-15e2 - 24e - 144 2e + 12

s0

15

Page 148

0

= =

0 6.75

6.70

Option (D) is correct. - 3 - 1 x1 x1 1 =x G = = 2 0 G=x G + = 0 Gu 2 2 x1 1 and Y = [1 0]= G + = Gu x2 2 -3 -1 1 Here , B = = G and C = [1 0] A == G 2 0 0 The controllability matrix is 1 -3 QC = [B AB ] = = 0 2G

s (s + 27) s + 29s + 6

or

which gives

Thus observable

6.72

6.73

6.74

6.76

e

s 0 1 0 s-1 0 -= == G G 0 s 0 1 0 s - 1G

1 0 (s - 1) s-1 1 = = = 0 (s - 1)G > 0 (s - 1) 2 -1

= L [(sI - A)]

K =- s (s2 + 5s2 + 6s) dK =- (3s2 + 10s + 6) = 0 ds s = - 10 ! 100 - 72 =- 0.784, - 2.548 6

The location of poles on s - plane is

Option (B) is correct.

At

1 + G (s) H (s) = 0 K 1+ =0 s (s + 2)( s + 3)

Since breakpoint must lie on root locus so s =- 0.748 is possible.

det Q0 ! 0

(sI - A)

h

Option (D) is correct. We have or

Q0 = [CT AT CT ] 1 -3 == !0 0 - 1G

-1

s + 27 s

Thus controllable

det QC ! 0 The observability matrix is

(sI - A) = =

^

1 + 29s + s62

2

We have

6.71

^ s +s27 h = 1 - ^ -s3 - 24s - s2 h + -s2 . -s3

-1

et 0 == G 0 et

Option (A) is correct.

0 1 s-1

Option (A) is correct. The given bode plot is shown below

A I D

H

O N

.in

co ia.

d Z = P-N no . w N " Net encirclement of (- 1 + j0) by Nyquist plot, ww P " Number of open loop poles in right hand side of s - plane Z " Number of closed loop poles in right hand side of s - plane Here N = 1 and P = 1 Thus Z =0 Hence there are no roots on RH of s -plane and system is always stable.

At w = 0.1 change in slope is + 60 dB " 3 zeroes at w = 0.1 At w = 10 change in slope is - 40 dB " 2 poles at w = 10 At w = 100 change in slope is - 20 dB " 1 poles at w = 100 K ( 0s.1 + 1) 3 Thus T (s) = s s ( 10 + 1) 2 ( 100 + 1) Now 20 log10 K = 20 or K = 10 10 ( 0s.1 + 1) 3 108 (s + 0.1) 3 Thus T (s) = s = s ( 10 + 1) 2 ( 100 + 1) (s + 10) 2 (s + 100)

Option (C) is correct. PD Controller may accentuate noise at higher frequency. It does not effect the type of system and it increases the damping. It also reduce the maximum overshoot.

Option (B) is correct. The characteristics equation is

6.77

s2 + 4s + 4 = 0 Comparing with

Option (D) is correct. Mason Gain Formula T (s) =

s2 + 2xwn + wn2 = 0 we get 2xwn = 4 and wn2 = 4 Thus x =1 ts = 4 = 4 = 2 1#2 xwn

Spk 3 k 3

In given SFG there is only forward path and 3 possible loop. p1 = 1 31 = 1 + 3 + 24 = s + 27 s s s L1 = - 2 , L2 = - 24 and L3 = - 3 s s s where L1 and L3 are non-touching C (s) This R (s) p1 3 1 = 1 - (loop gain) + pair of non - touching loops

6.78 6.79

Critically damped

Option (B) is correct. Option (C) is correct. We have 1 xo1 =xo G = =1 2 1 A == 1 s (sI - A) = = 0

0 x1 x1 (0) 1 and = == G G G G = 1 x2 x2 (0) 0 0 1G 0 1 0 s-1 0 -= G = = G s 1 1 - 1 s - 1G

GATE Electronics & Communication by RK Kanodia (Now in 3 Volumes) Purchase Online from our portal www.nodia.co.in and get maximum available discount

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia 1 (s - 1) 0 s-1 1 (sI - A) = > H = > +1 (s - 1) 2 + 1 (s - 1) (s - 1) t e 0 L-1 [(sI - A) -1] = eAt = = t t G te e et 0 1 et x (t) = eAt # [x (t0)] = = t t G= G = = t G te e 0 te

0

-1

6.80

6.83

s5

2

4

2

s4

1

2

1

s3

0

0

0

s2 s1 s0 Given x = 0.5

6.87

A I D

O N

or or

6.85

Option (B) is correct. Routh table is shown below. Here all element in 3rd row are zero, so system is marginal stable.

Option (B) is correct. The open loop transfer function is 1 G (s) H (s) = 2 s (s + s + 1) Option (B) is correct. Substituting s = jw we have Any point on real axis lies on the root locus if total number of poles 1 G (jw) H (jw) = and zeros to the right of that point is odd. Here s =- 1.5 does not jw (- w2 + jw + 1) lie on real axis because there are total two poles and zeros (0 and +G (jw) H (jw) =- p - tan-1 w 2 - 1) to the right of s =- 1.5 . 2 (1 - w ) Option (D) is correct. The frequency at which phase becomes - 180c, is called phase From the expression of OLTF it may be easily see that the maximum crossover frequency. magnitude is 0.5 and does not become 1 at any frequency. Thus gain wf Thus - 180 =- 90 - tan-1 cross over frequency does not exist. When gain cross over frequency 1 - wf2 does not exist, the phase margin is infinite. wf or - 90 =- tan-1 Option (D) is correct. 1 - wf2 n i . .coro 1 - w2f = 0 We have ...(i) xo (t) =- 2x (t) + 2u (t) a i d wf = 1 rad/sec Taking Laplace transform we get .no w The gain margin at this frequency wf = 1 is sX (s) =- 2X (s) + 2U (s) ww GM =- 20 log10 G (jwf) H (jwf) or (s + 2) X (s) = 2U (s) 2U (s) = 20 log10 (wf (1 - w2f) 2 + w2f or X (s) = (s + 2) =- 20 log 1 = 0 Now

6.84

6.86

ks2 + s + 6 = 0 s2 + 1 s + 6 = 0 K K

Comparing with s2 + 2xwn s + wn2 = 0 we have we get 2xwn = 1 and wn2 = 6 K K or 2 # 0.5 # 6 Kw = 1 K 6 = 1 & K =1 or K 6 K2

6.82

pole consideration is at s =- 1. Thus 1 =1 and Ts = 4 = 4 sec T T

H

Option (C) is correct. The characteristics equation is or

6.81

2

1 s-1

Page 149

y (t) = 0.5x (t) Y (s) = 0.5X (s) 0.5 # 2U (s) Y (s) = s+2 Y (s) 1 = (s + 2) U (s)

6.88

Z = P-N N " Net encirclement of (- 1 + j0) by Nyquist plot, P " Number of open loop poles in right had side of s - plane Z " Number of closed loop poles in right hand side of s - plane Here N = 0 (1 encirclement in CW direction and other in CCW) and P = 0 Thus Z = 0 Hence there are no roots on RH of s - plane.

Option (D) is correct. From Mason gain formula we can write transfer function as K Y (s) K s = = 3 R (s) 1 - ( s + -sK ) s - 3 (3 - K) For system to be stable (3 - K) < 0 i.e. K > 3 Option (B) is correct. The characteristics equation is (s + 1)( s + 100) = 0 s2 + 101s + 100 = 0 Comparing with s2 + 2xwn + wn2 = 0 we get 2xwn = 101 and wn2 = 100 Thus x = 101 20

6.89

Overdamped

For overdamped system settling time can be determined by the dominant pole of the closed loop system. In given system dominant

Option (A) is correct.

6.90

Option (D) is correct. Take off point is moved after G2 as shown below

Option (C) is correct. The characteristics equation is s2 + 2s + 2 = 0

GATE GUIDE and GATE CLOUD by RK Kanodia & Ashish Murolia GATE GUIDEs and CLOUDs are subjectwise books exclusive for GATE EC Examination. Download a sample chapter at

www.nodia.co.in

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

Page 150

Comparing with s2 + 2xwn + wn2 = 0 we get 2xwn = 2 and wn2 = 2

The Characteristic equation is 1 + G (s) H (s) = 0 K (s - 2) 1+ (s - 2) = 0 (s + 2) 2 or (s + 2) 2 + K (s - 2) 2 = 0 or (1 + K) s2 + 4 (1 - K) s + 4K + 4 = 0 Routh Table is shown below. For System to be stable 1 + k > 0 , and 4 + 4k > 0 and 4 - 4k > 0 . This gives - 1 < K < 1 As per question for 0 # K < 1

wn =

2 and x = 1 2 Since x < 1 thus system is under damped 6.91

6.92

Option (D) is correct. If roots of characteristics equation lie on negative axis at different positions (i.e. unequal), then system response is over damped. From the root locus diagram we see that for 0 < K < 1, the roots are on imaginary axis and for 1 < K < 5 roots are on complex plain. For K > 5 roots are again on imaginary axis. Thus system is over damped for 0 # K < 1 and K > 5 . Option (C) is correct. From SFG we have I1 (s) I2 (s) V0 (s) Now applying

= G1 Vi (s) + HI2 (s) = G2 I1 (s) = G3 I2 (s) KVL in given block diagram we have

...(1) ...(2) ...(3)

or

I1 (s) Z3 (S) = I2 (s)[ Z2 (s) + Z3 (s) + Z4 (s)] I1 (s) Z3 (s) Is (s) = Z3 (s) + Z2 (s) + Z4 (s)

Comparing (2) and (7) we have Z3 (s) G2 = Z3 (s) + Z2 (s) + Z4 (s) Comparing (1) and (6) we have Z3 (s) H = Z1 (s) + Z3 (s) 6.93

...(4) ...(5)

or

1 = s2 + 7s + 13 - 1 = s2 + 6s + 9 G (s) s+4 s+4 G (s) = 2 s + 4 s + 6s + 9

For DC gain s = 0 , thus Thus

G (0) = 4 9

Option (C) is correct. From the Block diagram transfer function is G (s) T (s) = 1 + G (s) H (s) K (s - 2) Where G (s) = (s + 2) and

H (s) = (s - 2)

s1

4 - 4k

0

s0

4 + 4k

IA

Thus a > 0 and aK > 3

...(6)

...(7)

in co.

.

ia od

n

. ww

w

6.97

Option (B) is correct. For unity negative feedback system the closed loop transfer function is G (s) , CLTF = G (s) " OL Gain = 2 s+4 1 + G (s) s + 7s + 13 2 1 + G (s) or = s + 7s + 13 G (s) s+4 or

6.94

N

4 + 4k

Option (B) is correct. The characteristics equation is s2 + as2 + ks + 3 = 0 The Routh Table is shown below For system to be stable a > 0 and aK - 3 > 0 a

D O

From (5) we have

1+k

Option (B) is correct. It is stable at all frequencies because for resistive network feedback factor is always less than unity. Hence overall gain decreases.

6.95

6.96

Vi (s) = I1 (s) Z1 (s) + [I1 (s) - I2 (s)] Z3 (s) 0 = [I2 (s) - I1 (s)] Z3 (s) + I2 (s) Z2 (s) + I2 (s) Z4 (s) From (4) we have or Vi (s) = I1 (s)[ Z1 (s) + Z3 (S)] - I2 (s) Z3 (S) Z3 (s) 1 or I1 (s) = Vi + I2 Z1 (s) + Z3 (s) Z1 (s) + Z3 (s)

s2

s3

1

K

s2

a

3

s1

aK - 3 a

0

s0

3

Option (B) is correct. Closed loop transfer function is given as T (s) = 2 9 s + 4s + 9 by comparing with standard form we get natural freq. wA2 = 9 wn = 3 2xwn = 4 4 = 2/3 2#3 for second order system the setting time for 2-percent band is given by 4 =4 =2 ts = 4 = xwn 3 # 2/3 2 x =

damping factor

6.98

Option (D) is correct. Given loop transfer function is G (s) H (s) =

2 s (s + 1)

2 jw (jw + 1) Phase cross over frequency can be calculated as G (jw) H (jw) =

So here

f (w) at w = w =- 180c f (w) =- 90c - tan-1 (w) p

- 90c - tan-1 (wp) =- 180c tan-1 (wp) = 90c

GATE Electronics & Communication by RK Kanodia (Now in 3 Volumes) Purchase Online from our portal www.nodia.co.in and get maximum available discount

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

Page 151

wp = 3

for unity feed back H (s) = 1

Gain margin 20 log 10 =

R (s) = 12 (unit Ramp) s 1 so E (s) = lim 12 s " 0 s 1 + G (s) n a1 sn - 1 + .... + an - 2 s2 = lim 12 s + n s"0 s s + a1 sn - 1 + .... + an = an - 2 an Here input

1 at w = wp G (jw) H (jw) G G.M. = 20 log 10 e

1 G (jw) H (jwp) o

2 =0 wp w2p + 1 G.M. = 20 log 10 b 1 l = 3 0

G (jwp) H (jwp) = so 6.99

6.103 6.104

Option (A) is correct. A ==

Here

Option (B) is correct.

0 1 0 , B = = G and C = [1 1] G 2 -3 1

6.105 6.106

The controllability matrix is QC = [B AB ] = =

Option (A) is correct. Option (A) is correct. By applying Routh’s criteria

0 1 1 - 3G

s3 + 5s2 + 7s + 3 = 0 Thus controllable

det QC ! 0 The observability matrix is 1 2 !0 Q0 = [CT AT CT ] = = 1 - 2G

Thus observable

det Q0 ! 0 6.100

G (s) H (s) = 2 3 s (s + 1) G (jw) H (jw) =

or

2 3 jw (jw + 1)

G (jw) H (jw) at w = w = 1 g

2 3 =1 w w2 + 1 12 = w2 (w2 + 1) w4 + w2 - 12 = 0 (w2 + 4) (w2 - 3) = 0 w2 = 3 and w2 =- 4 wg =

3

s1

7#5-3 5

s0

3

=

0

32 5

O N

6.108

H (s) = 1 (unity feedback)

R (s) = 1 s

3 f (w) at w = w =- 90 - tan-1 (wg)

6.109

g

Option (B) is correct. Option (C) is correct. Closed-loop transfer function is given by an - 1 s + an T (s) = n n-1 s + a1 s + ... + an - 1 s + an an - 1 s + an n n-1 + + ...an - 2 s2 s a 1s = an - 1 s + an 1+ n s + a1 sn - 1 + ...an - 2 s2 an - 1 s + an Thus G (s) H (s) = n s + a1 sn - 1 + ....an - 2 s2 For unity feed back H (s) = 1 an - 1 s + an Thus G (s) = n s + a1 sn - 1 + ....an - 2 s2 Steady state error is given by 1 E (s) = lim R (s) s"0 1 + G (s) H (s)

R (s) = input

Where

g

6.102

5

Option (A) is correct. Open loop transfer function is n K o.i c G (s) = . a i s ( s + 1) d o n Steady state error w. w sR (s) w E (s) = lim s " 0 1 + G (s) H (s)

=- 90 - tan-1 3 =- 90 - 60 =- 150 Phase margin = 180 + f (w) at w = w = 180 - 150 = 30c 6.101

s2

Option (A) is correct. Techometer acts like a differentiator so its transfer function is of the form ks .

w1, w2 = ! 3

which gives

7

A I D 6.107

Gain cross over frequency

or

1

There is no sign change in the first column. Thus there is no root lying in the left-half plane.

Option (D) is correct. we have

s3

s1 s (s + 1) s so = lim 2 =0 E (s) = lim s"0 s"0 s + s + K K 1+ s (s + 1) Option (B) is correct. Fig given below shows a unit impulse input given to a zero-order hold circuit which holds the input signal for a duration T & therefore, the output is a unit step function till duration T .

h (t) = u (t) - u (t - T) Taking Laplace transform we have H (s) = 1 - 1 e-sT = 1 61 - e-sT @ s s s 6.110

Option (C) is correct. Phase margin = 180c + qg where qg = value of phase at gain crossover frequency.

GATE GUIDE and GATE CLOUD by RK Kanodia & Ashish Murolia GATE GUIDEs and CLOUDs are subjectwise books exclusive for GATE EC Examination. Download a sample chapter at

www.nodia.co.in

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

Here so 6.111

6.112

Page 152

Xo2 = dx2 = 0 + x2 + m dt x1 y = [1 1] > H = x1 + x2 x2 dy = dx1 + dx2 dt dt dt dy = x1 + m dt

qg =- 125c P.M = 180c - 125c = 55c

Option (B) is correct. Open loop transfer function is given by K (1 + 0.5s) G (s) H (s) = s (1 + s) (1 + 2s) Close looped system is of type 1. It must be noted that type of the system is defined as no. of poles lies at origin in OLTF. lying Option (D) is correct. Transfer function of the phase lead controller is 1 + (3Tw) j T.F = 1 + 3Ts = 1+s 1 + (Tw) j Phase is

dy dt

= x1 (0) + m (0) t=0

= 1+0 = 0

f (w) = tan-1 (3Tw) - tan-1 (Tw) w f (w) = tan-1 ; 3Tw - T 1 + 3T 2 w2 E f (w) = tan-1 ; 2Tw2 2 E 1 + 3T w For maximum value of phase df (w) =0 dw 1 = 3T 2 w2 Tw = 1 3 So maximum phase is or

6.113

6.114

A I D

fmax = tan-1 ; 2Tw2 2 E at Tw = 1 1 + 3T w 3 R V 1 S 2 W .in 3 W = tan-1 1 = 30c o c = tan-1 S . ; 3E SS1 + 3 # 1 WW dia o 3 .n T X w Option (A) is correct. ww G (jw) H (jw) enclose the (- 1, 0) point so here G (jwp) H (jwp) > 1 wp = Phase cross over frequency 1 Gain Margin = 20 log 10 G (jwp) H (jwp) so gain margin will be less than zero.

O N

Option (B) is correct. The denominator of Transfer function is called the characteristic equation of the system. so here characteristic equation is (s + 1) 2 (s + 2) = 0

6.115

6.116

Option (C) is correct. In synchro error detector, output voltage is proportional to [w (t)], where w (t) is the rotor velocity so here n = 1 Option (C) is correct. By masson’s gain formulae / Dk Pk y = x D Forward path gain

so gain 6.117

P1 = 5 # 2 # 1 = 10 D = 1 - (2 # - 2) = 1 + 4 = 5 D1 = 1 y = 10 # 1 = 2 x 5

Option (C) is correct. By given matrix equations we can have Xo1 = dx1 = x1 - x2 + 0 dt

GATE Electronics & Communication by RK Kanodia (Now in 3 Volumes) Purchase Online from our portal www.nodia.co.in and get maximum available discount